Update to v0.2:

Changed: more precise statement of the bound on the large x decay of u (Theorem 1.2)

  Added: more details in the proof of the large x decay of u (Theorem 1.2)

  Fixed: miscellaneous typos.

Carlen_Jauslin_Lieb_2020.tex

@@ -125,10 +125,10 @@ Along with the computation of the low density energy of the simple equation in o
   for $N$ particles in a cubic box of  finite volume $V$ with periodic boundary conditions.
   The ground state eigenfunction $\psi_0$ is unique and non-negative, as can be shown using the Perron-Frobenius theorem, and thus we may normalize $\psi_0$ to obtain a probability measure.  This is not the usual probability measure associated to a quantum state, which would be quadratic in the wave function, but since $\psi_0$ is non-negative and integrable ($\|\psi_0\|_1 \leqslant V^{1/2}\|\psi_0\|_2$), we may use it directly to define a probability measure, and this is the starting point of \cite{Li63}.  Because particles interact pairwise,  the ground state energy and other observables can be calculated in terms of the two-point correlation function associated to this probability measure:
 \begin{equation}
-  g_N(x_1-x_2) := \lim_{N,V\to\infty, N/V = \rho}\frac{V^2\int dx_3\cdots dx_N\ \psi_0(x_1,x_2,x_3,\dots,x_N)}{\int dy_1\cdots dy_N\ \psi_0(y_1,\dots,y_N )}
+  g(x_1-x_2) := \lim_{N,V\to\infty, N/V = \rho}\frac{V^2\int dx_3\cdots dx_N\ \psi_0(x_1,x_2,x_3,\dots,x_N)}{\int dy_1\cdots dy_N\ \psi_0(y_1,\dots,y_N )}
 \end{equation}
-  In \cite{Li63}, under a few physically motivated approximations, in the thermodynamic limit, in which the number of particles $N$ and the volume of the gas $V$ are taken to infinity, with $\rho:=\frac NV$ fixed, an equation for the limiting two-point correlation function $g_{\infty}$ is derived.
+  In \cite{Li63}, under a few physically motivated approximations, in the thermodynamic limit, in which the number of particles $N$ and the volume of the gas $V$ are taken to infinity, with $\rho:=\frac NV$ fixed, an equation for the limiting two-point correlation function $g$ is derived.
-  The function $u(x)$ in \eqref{simp} is then defined as $u(x):=1- g_{\infty}(x)$. Note that since by definition $g_{\infty}(x) \geqslant 0$, $u(x) \leqslant 1$. 
+  The function $u(x)$ in \eqref{simp} is then defined as $u(x):=1- g(x)$. Note that since by definition $g(x) \geqslant 0$, $u(x) \leqslant 1$. 
 
 
   Because the expected values in the ground state of  many physical observables can be calculated in terms of $g$, any method for computing $g$ that bypasses directly solving the $N$-body Schr\"odinger equation for the Hamiltonian  \eqref{ham0} provides an effective means for the computation of these  values, and this motivates the study of  the simple equation system\-~(\ref{simp}). Indeed, the ground state energy per particle is given in terms of $g$ by the second equation in \eqref{simp}.  There is so far no rigorous derivation of \eqref{simp} from the $N$-body Schr\"odinger equation,  and hence there is no mathematical 
@@ -325,7 +325,7 @@ We shall also need various $L^p$ bounds on $u'$, and for these we need a detaile
   If $(1+ |x|^4)v(x)\in L^1(\mathbb R^3)\cap L^2(\mathbb R^3)$, then
   \begin{equation}
     \rho u(x)=
-    \frac{\sqrt{2+\beta}}{2\pi^2\sqrt{e}}\frac1{1+|x|^{4}}
+    \frac{\sqrt{2+\beta}}{2\pi^2\sqrt{e}}\frac1{|x|^{4}}
     +
     R(x)
     \label{udecay}
@@ -336,18 +336,13 @@ We shall also need various $L^p$ bounds on $u'$, and for these we need a detaile
     ,
   \end{equation}
   and  where $|x|^4R(x)$ is in $L^2(\mathbb R^3)\cap L^\infty(\R^3)$, uniformly in $e$ on all compact sets.
- Moreover,  there is a constant $C$ independent of $e$ and $\rho$ such that for all $x$
+ Moreover, for every $\rho_0>0$, there is a constant $C$ that only depends on $\rho_0$ such that for all $x$, for all $\rho<\rho_0$,
   \begin{equation}\label{fourdecay}
- u(x) \leqslant C \rho^{-1}e^{-1/2}\frac{1}{1 + |x|^4}\ . 
+   u(x) \leqslant \min\left\{1,\frac C{\rho e^{\frac12}|x|^4}\right\}\ .
   \end{equation}
 \end{theorem}
  
 
-\begin{remark}
-  Theorem\-~\ref{theo:pointwise} actually holds with the condition that $|x|^4 v\in L^1(\mathbb R^3)$, without necessarily being in $L^2(\mathbb R^3)$, together with the condition $v\in L^1(\R^3)\cap L^2(\R^3)$, but the proof of that is a bit more involved, and will not be presented here.  
-\end{remark}
- 
-
 The next two theorems concern the monotonicity of $\rho\mapsto e(\rho)$ and convexity if $\rho\mapsto\rho e(\rho)$.
 These were conjectured in\-~\cite{CJL20}, and here, we prove them for small density $\rho$ (and, in the case of the monotonicity, also for large density).
 
@@ -600,7 +595,7 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
 \begin{remark}
   As stated at the very beginning of the paper, we assume that $v$ is spherically symmetric.
   This is, however, used very little in the proofs.
-  In fact, the only theorem which relies on the spherical symmetry is theorem\-~\ref{theo:pointwise}.
+  In fact, the only theorem that relies on the spherical symmetry is theorem\-~\ref{theo:pointwise}.
   We believe it should still hold (provided the decay constant in\-~(\ref{udecay}) is suitably adapted) without the spherical symmetry.
   In this case, the other theorems would not require the spherical symmetry.
 \end{remark}
@@ -665,6 +660,7 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
     U_1(x)=\frac{e^{\frac32}}\pi e^{-2\sqrt e|x|}
     -\frac1\pi \left(\delta(x)+\frac{(\beta+1)e}\pi\frac{e^{-2\sqrt e|x|}}{|x|}\right)
     \ast f_1\ast f_2
+    \label{U1}
   \end{equation}
   where
   \begin{equation}
@@ -676,11 +672,13 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
     =\frac{e}{\pi|x|}  \int_0^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2} dt\ ,
   \end{equation}
   now, for all $T>0$,
-  \begin{eqnarray*}
+  \begin{eqnarray}
-  \int_0^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2}\ dt &=& \int_0^{T} e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2}+\int_{T}^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2}\ dt\\
+  \int_0^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2}\ dt &=& \int_0^{T} e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2}+\int_{T}^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)}  t^{-1/2}\ dt
-  &\leqslant& \int_0^{T} e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)}  t^{-1/2}+\int_{T}^\infty e^{-\sqrt{ t}(2\sqrt{e}|x|)}  t^{-1/2}\ dt\\
+  \nonumber\\
-  &=& 2 T^{1/2} e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)}  +  \frac{1}{\sqrt{e}|x|}e^{-\sqrt{T}(2\sqrt{e}|x|)}\ .
+  &\leqslant&
-  \end{eqnarray*}
+  2 T^{1/2} e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)}  +  \frac{1}{\sqrt{e}|x|}e^{-\sqrt{T}(2\sqrt{e}|x|)}\ .
+  \label{boundf2}
+  \end{eqnarray}
   Choosing $T = 2+\beta$, we see that for large $(2\sqrt{e}|x|)$, $0 \leqslant f_2(x) \leqslant C e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)}$.
   Furthermore,
   \begin{equation}
@@ -689,14 +687,16 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
     \frac{e^{\frac32}}{\pi^3}\frac1{|x|^2}\ast g
     ,\quad
     g(x)=\frac{(1-\sqrt e|x|)e^{-(2\sqrt e)|x|}}{|x|}
+    \label{g}
   \end{equation}
   Using 
   \begin{equation}
-  \frac{1}{|x-y|^2} = \frac{1}{1+|x|^2} +  \frac{1-|y|^2 + 2x\cdot y}{ (1+|x|^2)|x-y|^2}
+  \frac{1}{|x-y|^2} = \frac{1}{|x|^2} +  \frac{-|y|^2 + 2x\cdot y}{|x|^2|x-y|^2}
   \end{equation}
   twice and the fact that $g(y)$ is even, integrates to zero, and $\int y g(y)\ dy=0$,
   \begin{equation}
-  f_1(x) = \frac{1}{(1+|x|^2)^2}  \frac{e^{\frac32}}{\pi^3}\left( - \int_{\R^3} |y|^2 g(y){\rm d} y  +   \int_{\R^3}  \frac{(1-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}g(y) {\rm d} y\right)
+  f_1(x) = \frac{1}{|x|^4}  \frac{e^{\frac32}}{\pi^3}\left( - \int_{\R^3} |y|^2 g(y){\rm d} y  +   \int_{\R^3}  \frac{(-|y|^2 + 2x\cdot y)^2}{|x-y|^2}g(y) {\rm d} y\right)
+  \label{f1}
   \end{equation}
   We compute
   $
@@ -708,7 +708,58 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
   Therefore,
   \begin{equation}
   \lim_{|x|\to \infty} |x|^4 f_1(x) =  -\frac{1}{2\pi^2\sqrt e} \qquad{\rm and}\qquad   \lim_{|x|\to \infty} |x|^4 U_1(x)  =  \frac{1}{2\pi^2\sqrt e}\sqrt{2+\beta}\ .
+  \label{U1decay}
   \end{equation}
+  \bigskip
+
+  \indent
+  We now turn to an upper bound of $U_1$.
+  First of all, if $|x|\leqslant\frac1{\sqrt e}$, then by\-~(\ref{g}) and\-~(\ref{f1}),
+  \begin{equation}
+    f_1(x)\geqslant 0
+  \end{equation}
+  and if $|x|>\frac1{\sqrt e}$, then
+  \begin{equation}
+     f_1(x)\geqslant
+     -\frac{1}{|x|^4}  \frac{e^{2}}{\pi^3}\int_{\mathbb R^3}  \frac{(-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}e^{-(2\sqrt e)|y|} {\rm d} y
+     .
+  \end{equation}
+  We split the integral into two parts: $|y-x|>|x|$ and $|y-x|<|x|$.
+  We have, (recalling $|x|>\frac1{\sqrt e}$),
+  \begin{equation}
+     \int_{|y-x|>|x|}  \frac{(-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}e^{-(2\sqrt e)|y|} {\rm d} y
+     \leqslant
+     e^{-\frac52}C
+  \end{equation}
+  for some constant $C$ (we use a notation where the constant $C$ may change from one line to the next).
+  Now,
+  \begin{equation}
+     \int_{|y-x|<|x|}  \frac{(-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}e^{-(2\sqrt e)|y|} {\rm d} y
+     \leqslant
+     e^{-\sqrt e|x|}\int_{|y-x|<|x|}  \frac{(|y|^2 + 2|x||y|)^2}{ |x-y|^2} {\rm d} y
+     \leqslant
+     |x|^5e^{-\sqrt e|x|}C
+     .
+  \end{equation}
+  Therefore, for all $x$,
+  \begin{equation}
+    f_1(x)\geqslant
+    -\frac{1}{|x|^4}C(e^{-\frac12}+e^2|x|^4e^{-\sqrt e|x|})
+    .
+  \end{equation}
+  Finally, by use\-~(\ref{boundf2}),
+  \begin{equation}
+    |x|^4\left(\delta(x)+\frac{(\beta+1)e}{\pi}\frac{e^{-2\sqrt e|x|}}{|x|}\right)\ast f_1\ast f_2(x)\geqslant
+    -Ce^{-\frac12}
+    .
+  \end{equation}
+  All in all, by\-~(\ref{U1}), (since $|x|^4e^{\frac32}e^{-2\sqrt e|x|}<Ce^{-\frac12}$)
+  \begin{equation}
+    |x|^4U_1(x)\leqslant Ce^{-\frac12}
+    .
+    \label{boundU1}
+  \end{equation}
+  \bigskip
 
   \point
   We now show that $\Delta^2\widehat U_2$ is integrable and square-integrable.
@@ -716,6 +767,7 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
   \begin{equation}
     16e^2\Delta^2\equiv\partial_\kappa^4+\frac4\kappa\partial_\kappa^3
     .
+    \label{D2}
   \end{equation}
   We have, by the Leibniz rule,
   \begin{equation}
@@ -739,6 +791,7 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
   for some family of constants $c_{l_1,\cdots,l_p}^{(p,n)}$ which can easily be computed explicitly, but this is not needed.
   Now, since $S\geqslant 0$, $\frac\rho{1e}|\widehat S|\leqslant 1$, so $|\zeta_1|\leqslant\frac12$ and $\zeta_1=\frac12$ if and only if $\kappa=0$.
   Therefore, $\widehat U_2$ is bounded when $\kappa$ is away from 0, so it suffices to show that $\Delta^2\widehat U_2$ is integrable and square integrable at infinity and at 0.
+  \bigskip
    
 
   \subpoint We first consider the behavior at infinity, and assume that $\kappa$ is sufficiently large.
@@ -746,7 +799,7 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
   To prove the corresponding claim for $\zeta_1$, we use the fact that $|x|^4 v$ square integrable, which implies that $\widehat S$ is as well.
   Therefore, by\-~(\ref{zetas}) for $0\leqslant n\leqslant 4$, $\kappa^2\partial_\kappa^n\zeta_1$ is integrable at infinity, and, therefore, square-integrable at infinity.
   Furthermore, by\-~(\ref{zetas}), $\zeta_1<\frac12-\epsilon$ for large $\kappa$, and $\partial^n\zeta_1$ is bounded, so $\partial_\kappa^{n-i}(\kappa^2+1)\partial_\kappa^i(1-\sqrt{1-2\zeta_1})$ is integrable and square integrable.
-   
+  \bigskip
 
   \subpoint As $\kappa\to0$
   \begin{equation}
@@ -787,7 +840,7 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
     =O(\kappa^{1-i})
     .
   \end{equation}
-  Thus, by\-~(\ref{leibnitz}),
+  Thus, by\-~(\ref{leibnitz}), as $\kappa\to0$,
   \begin{equation}
     |\partial_\kappa^4\widehat U_2|=O(\kappa^{-1})
     ,\quad
@@ -795,7 +848,15 @@ This provides  the control on $\|u'\|_q$ that we need to prove the theorem on co
     .
   \end{equation}
   Thus, $\Delta^2\widehat U_2$ is integrable and square integrable.
-  And since the $O(\cdot)$ hold uniformly in $e$ on all compact sets, $U_2|x|^4$ is bounded and square integrable uniformly in $e$ on all compact sets.
+  And since the $O(\cdot)$ hold uniformly in $e$ on all compact sets, by\-~(\ref{D2}),
+  \begin{equation}
+    |x|^4U_2(x)
+    \leqslant
+    \frac{8e^{\frac32}}{16e^2}\int \left(\partial_{|k|}^4+\frac4{|k|}\partial_{|k|}^3\right)\hat U_2(|k|)\ dk
+    \leqslant\frac C{\sqrt e}
+    .
+  \end{equation}
+  This along with\-~(\ref{U1decay}) and\-~(\ref{boundU1}) implies\-~(\ref{udecay}) and\-~(\ref{fourdecay}).
 \qed
 
 
@@ -1849,7 +1910,7 @@ $$
 $$
 \end{proof}
 
-It now follows that with $e_\star$ define as in  Theorem~\ref{Mon}, on account of the bound on $\rho'$ proved there, and on account of Theorem~\ref{theo:pointwise} that there is a constant independent of $e$ such that for all $e\leqslant e_\star$,
+It now follows that with $e_\star$ defined as in  Theorem~\ref{Mon}, on account of the bound on $\rho'$ proved there, and on account of Theorem~\ref{theo:pointwise} that there is a constant independent of $e$ such that for all $e\leqslant e_\star$,
 \begin{equation}\label{varphipointwise}
 |\varphi(x)| \leqslant Ce^{-3/2}(1 + |x|^4)^{-1}\ .
 \end{equation}

Changelog

@@ -1,3 +1,12 @@
+v0.2:
+
+  * Changed: more precise statement of the bound on the large x decay of u (Theorem 1.2)
+
+  * Added: more details in the proof of the large x decay of u (Theorem 1.2)
+
+  * Fixed: miscellaneous typos.
+
+
 v0.1:
 
   * Added: references to [Nelson, 1964] and [Reed, Simon, 1975].